Answer:
4+3x
It's pretty straight forward, you have the number 4 that is added to the rest of the expression because sum is addition, and the product of 2 numbers is multiplication so it would make 3(x) or 3x, which results in 4+3x.
Tables are created by substitute a set of input values into the function to create outputs. The required table is as shown below
<em>x | y</em>
<em>0 -3 </em>
<em>1 -2.5</em>
<em>2 -2 </em>
<h3>Tables and values</h3>
Tables are created by substitute a set of input values into the function to create outputs
Using x = 0, 1 and 2 as the input values
Given the function
y = 1/2x - 3
If x = 0
y = 1/2(0) - 3
y = -3
If x = 1
y = 1/2(1) - 3
y = -2.5
If x = 2
y = 1/2(2) - 3
y = -2
Hence the required table is as shown below
x | y
0 -3
1 -2.5
2 -2
Learn more on tables and values here: brainly.com/question/12151322
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Answer:
You will pay $42.50
Step-by-step explanation:
You will pay 50-7.5 which gives you 42.50
Answer:
a=29/42
Step-by-step explanation:
6a-1÷7=4
first write the division as a fraction
6a-1/7=4
add 1/7 to each side
6a=4+1/7
which is equal to
6a=29/7
divide both sides by 6
a=29/42
Option D. D has the matrix of constants [[12], [11], [4]].
Step-by-step explanation:
Step 1:
With the given equations, we can form matrices to represent them.
The coefficients of x, y, and z form a matrix of order 3 ×3, the variables x, y, and z form a matrix of order 1 ×3 and the constants form a matrix of order 1 ×3.
Step 2:
The linear system A is represented as
![\left[\begin{array}{ccc}12&1&1\\1&-11&0\\1&-1&4\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D12%261%261%5C%5C1%26-11%260%5C%5C1%26-1%264%5Cend%7Barray%7D%5Cright%5D)
![\left[\begin{array}{ccc}x\\y\\z\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Dx%5C%5Cy%5C%5Cz%5Cend%7Barray%7D%5Cright%5D)
![= \left[\begin{array}{ccc}26\\17\\23\end{array}\right]](https://tex.z-dn.net/?f=%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D26%5C%5C17%5C%5C23%5Cend%7Barray%7D%5Cright%5D)
.
Step 3:
The linear system B is represented as
![\left[\begin{array}{ccc}4&1&1\\1&-11&0\\1&-1&12\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D4%261%261%5C%5C1%26-11%260%5C%5C1%26-1%2612%5Cend%7Barray%7D%5Cright%5D)
![= \left[\begin{array}{ccc}23\\17\\26\end{array}\right]](https://tex.z-dn.net/?f=%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D23%5C%5C17%5C%5C26%5Cend%7Barray%7D%5Cright%5D)
.
Step 4:
The linear system C is represented as
![\left[\begin{array}{ccc}1&1&1\\1&-1&0\\1&-1&1\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%261%261%5C%5C1%26-1%260%5C%5C1%26-1%261%5Cend%7Barray%7D%5Cright%5D)
![= \left[\begin{array}{ccc}4\\11\\12\end{array}\right]](https://tex.z-dn.net/?f=%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D4%5C%5C11%5C%5C12%5Cend%7Barray%7D%5Cright%5D)
.
Step 5:
The linear system D is represented as
![= \left[\begin{array}{ccc}12\\11\\4\end{array}\right]](https://tex.z-dn.net/?f=%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D12%5C%5C11%5C%5C4%5Cend%7Barray%7D%5Cright%5D)
.
Step 6:
Of the four options, the linear system D has the matrix of constants [[12], [11], [4]]. So the answer is option D. D.
